Answer
The magnitude of the friction force on each hand and foot is 160 N.
Work Step by Step
The sum of the four friction forces is equal in magnitude to the weight of the gymnast. So,
$4~F_f = mg$
$F_f = \frac{mg}{4}$
$F_f = \frac{(65~kg)(9.80~m/s^2)}{4}$
$F_f = 160~N$
The magnitude of the friction force on each hand and foot is 160 N.